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BabaBlast [244]

3 years ago

12

If the mother of a child has the recessive gene for cystic fibrosis (Cc), an autosomal disorder, and the father does not (CC), w

hat is the probability that the child will get cystic fibrosis?
A. 0%
B. 25%
C. 50%
D. 75%

1 answer:

Arisa [49]3 years ago

7 0

Answer:A

Explanation:

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The number of cells in an average-sized adult human is on the order of 10^14 cells. Use this information, and the estimate that

KengaRu [80]

IF:
Number of cells: $C = 10^{14}$
DNA lenght: $D = 2m = 12.42 * 10^{-4} mi$
Distance from Earth to Sun: $ES=93000000 mi = 9.3* 10^{7}mi$

Then:
a) <span>Over how many miles would the total DNA from the average human stretch?
The answer is product of multiplication of the number of cells (C) and the DNA length (D):
Total DNA: </span> $T = C*D$
⇒ $T = 10^{14}*12.42* 10^{-4} mi$
⇒ $T = 12.42* 10^{14-4} mi$
⇒ $T = 12.42* 10^{10} mi$
The total DNA from the average human will stretch $12.42* 10^{10} mi$

b) How many times would the total DNA from the average human stretch from Earth to the Sun and back?
The answer is concluded from the ratio of the total DNA length (T) and the <em>twice </em>(because of stretch to the Sun and back, thus, <em>two directions</em>) of distance from <span>Earth to the Sun (ES) and :
Ratio: </span> $R= \frac{T}{2*ES}$
⇒ $R = \frac{12.42* 10^{10}mi }{2*(9.3* 10^{7}mi)}$
⇒ $R = \frac{12.42* 10^{10}}{18.6*10^{7}}$
⇒ $R = 0.6677* 10^{10-7}$
<span> ⇒ $R = 0.6677* 10^{3}$
</span><span> ⇒ [tex]R = 667.7
</span>Thus, 667,7 times will the <span>total DNA from the average human stretch from Earth to the Sun and back.</span>

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Which of the following correctly lists the kingdoms of life scientists currently use

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I'm not sure if these 6 kingdoms I listed are one of the"which of the following"options you asked for so I hope it helps.

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Parenchyma
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